Name:
Math 1060
Exam 2 Review
1. Fill in the following table:
θ (radians)
θ (degrees)
sin θ
cos θ
tan θ
0
0
0
1
0
π/3
60
5π/6
150
1/2
5π/3
300
π
√
√
3/2
3
1/2
√
√
− 3/2
− 3/3
− 3/2
1/2
√
− 3
180
0
−1
0
3π/2
270
−1
0
undefined
17π/2
1530
1
0
undefined
3π/4
135
√
√
√
2/2
− 2/2
−1
2. Fill in the following table. It may help to refer to your answers from problem 1.
θ (radians)
sec θ
csc θ
cot θ
0
1
undefined
undefined
π/3
2
√
2 3
3
5π/6
− 233
2
5π/3
2
− 233
− 3/3
π
−1
undefined
undefined
3π/2
undefined
−1
0
17π/2
undefined
1
0
√
√
√
3/3
√
− 3
√
Exam 2 Review
Math 1060
3. For each of the following functions:
• Find the amplitude, period, phase shift, and vertical shift
• Graph the function, including at least one period
(a) f (x) = 5 sin(3πx − π) + 1
a=5
b = 3π
2π
period =
b
2
=
3
amplitude = |a|
=5
And it looks a bit like this:
c=π
c
phase shift =
b
1
=
3
d=1
vertical shift = d
=1
6
4
2
-1
-0.5
0.5
1
-2
-4
(b) g(x) = −2 cos
3π
2 x+2
a = −2
amplitude = |a|
=2
And it looks a bit like this:
3π
2
2π
period =
b
4
=
3
b=
c = −2
c
b
−4
=
3π
phase shift =
d=0
vertical shift = d
=0
Exam 2 Review
Math 1060
2
1
-1
-2
1
2
-1
-2
(c) h(t) = − sin
x
2
+ π2
1
2
2π
period =
b
= 4π
a = −1
b=
amplitude = |a|
=1
And it looks a bit like this:
c=−
π
2
c
b
= −π
phase shift =
d=0
vertical shift = d
=0
1
0.5
-10
-5
5
10
-0.5
-1
4. Graph the following functions. It may help to find the period, phase shift, and vertical shift first.
(a) f (x) = tan(πx) + 1
Exam 2 Review
Math 1060
4
3
2
1
-1
-0.5
1
0.5
-1
-2
(b) f (x) = 2 csc
π
π
2x− 2
10
5
-3
-2
-1
1
2
3
4
5
-5
-10
5. Simplify the following:
(a) sin cos−1 (x)
1
t
x
This triangle is an illustration of the equation
cos(t) = x
Remember
that this equation means the same thing as t = cos−1 (x). The remaining side has length
√
opposite
1 − x2 . Since sin(t) = hypotenuse , we have that
√
sin cos−1 (x) = sin(t) = 1 − x2
(b) cot sin−1 (x)
Exam 2 Review
Math 1060
1
x
t
This triangle is an illustration of the equation
sin(t) = x
Remember
that this equation means the same thing as t = sin−1 (x). The remaining side has length
√
adjacent
2
1 − x . Since cot(t) = opposite , we have that
√
cot sin−1 (x) = cot(t) =
1 − x2
x
(c) sin−1 cos 3π
4
√
to be between − π2
= − 22 . So what is sin−1
√ and π2 . Since sin − 22 = − π4 , we have that
Well, ok, we know that cos
3π
4
√ 2
2
? Remember that the answer has
√ !
3π
2
π
= sin−1 −
sin−1 cos
=−
4
2
4
that
If you want to make sure that you really understand this type of problem, convince yourself
3π
π
sin−1 sin
=
4
4
6. A lighthouse is 2 miles from the closest point on a perfectly straight shore. The light from the lighthouse is rotating; call the direction it is pointing the angle θ, where θ = 0 radians means the light is
pointing straight toward the shore. Say d is the distance from the lighthouse to the point on the shore
where the light is shining.
The paragraph above will probably be extremely confusing until you get a picture drawn.
(a) Write down a function giving the distance d in terms of the angle θ and graph it. Think carefully
about the domain of this function before you graph it!
Here’s a nice diagram for you:
Exam 2 Review
Math 1060
2 mi
θ
d
It looks like
2
d
d cos(θ) = 2
cos(θ) =
d=
2
cos(θ)
You could also write it:
d = 2 sec(θ)
Here’s a graph:
20
15
10
5
-1.5
-1
-0.5
0.5
1
1.5
Notice that there are vertical asymptotes at − π2 and
problem.
π
2,
which is what you’d expect from the
(b) Write down a function giving the angle θ in terms of the distance d and graph it.
The first thing to notice is that the question is impossible. There’s no such function! If you’re,
for instance, 3 miles away from the lighthouse, you could be either on the left side or the right
side of that diagram. (On a review I’ll expect you to notice that it’s impossible, state that fact, and
then figure out how to give a decent answer. On a test I’ll phrase the question so that issue doesn’t
come up.)
Exam 2 Review
Math 1060
Ok. We can basically solve this problem by simply solving the equation above for θ. So that’s
θ = sec−1
θ
2
But let’s think carefully about what this function means. We’ve defined sec−1 (x) to be always
positive. That means that the graph of the function is this:
1.5
1
0.5
0
5
15
10
20
-0.5
-1
-1.5
But again, you should notice that this function does not tell us whether we are to the left or the
right of the lighthouse.
7. A boat is being pulled into shore by a rope. The rope is anchored to a winch on shore, which is 5 feet
above the deck of the boat. Say s is the length of the rope from the winch to the boat, and θ is the angle
of elevation from the boat to the winch.
Again, a picture will be instrumental. In this case, if you are having trouble, see problem 91 on page
350.
(a) Write θ as a function of s.
Here’s a lovely picture for you:
s
5’
θ
I’ll write this in two ways. Either is fine.
sin(θ) =
5
s
csc(θ) =
s
5
Solving for θ:
5
s
Here’s a graph of θ = sin−1 5s = csc−1 5s :
θ = sin−1
θ = csc−1
s
5
Exam 2 Review
Math 1060
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
10
20
30
40
50
When s = 5 feet, the boat is right under the dock, so θ = π2 . As s gets larger and larger, θ
gets closer and closer to 0, because the rope is getting more and more horizontal. Notice that the
domain is [5, ∞].
(b) Write s as a function of θ.
The simplest way to get this is to just solve the function above for θ. Starting from the same
picture we have:
sin(θ) =
5
s
csc(θ) =
s
5
Solving for s:
s=
Here’s a graph of θ =
5
sin(θ)
5
sin(θ)
s = 5 csc(θ)
= 5 csc(θ):
50
40
30
20
10
1
0.5
h
1.5
i
The domain, of course, is 0, π2 .
(c) Find θ when s = 10 feet.
Let’s just plug in:
5
1
π
= sin−1
=
10
2
6
You should be able to figure that out without recourse to a calculator.
θ = sin−1